The first proofs of the Prime Number Theorem, by Hadamard and de la Vallée Poussin in 1896, were long and technical: their engine was a delicate zero-free region — a curved sliver to the left of the line \Re(s)=1 inside which \zeta was shown to have no zeros, wrung out of intricate estimates. For most of a century that seemed to be the price of admission.

Then, in 1980, Donald J. Newman found a proof so short it fits on a postcard. In 1997 Don Zagier wrote it up in a famous three-page article, "Newman's short proof of the prime number theorem", and it has been the way graduate courses teach the PNT ever since. The magic is that it needs almost nothing about the zeros of \zeta — only that there are none exactly on the line \Re(s)=1. No zero-free region, no error term, no fuss: just three clean ingredients bolted together by one slick contour integral. This page walks through that assembly.

The setup: a prime-powered Dirichlet series

We work not with \pi(x) directly but with Chebyshev's weighted count, the first Chebyshev function

which sums \log p over primes p \le x. Proving \theta(x) \sim x is equivalent to the PNT (a short partial-summation argument transfers it to \pi(x)\sim x/\ln x). The analytic object that "sees" \theta is the Dirichlet series

a prime-only cousin of the zeta sum. A Mellin transform ties the two together exactly: for \Re(s) > 1,

So \Phi is a smoothed-out Laplace/Mellin picture of \theta; controlling \Phi near s=1 is what will control the growth of \theta.

How Φ relates to −ζ′/ζ

Recall the logarithmic derivative of zeta, whose Dirichlet series is the von Mangoldt sum over all prime powers:

The k=1 slice of that double sum is exactly \Phi(s). The leftover higher powers,

converge absolutely for \Re(s) > \tfrac12 and so define a holomorphic function well past the line we care about. In other words, \Phi(s) and -\zeta'(s)/\zeta(s) differ only by something harmless near \Re(s)=1. Since \zeta has a simple pole at s=1 (giving -\zeta'/\zeta a pole with residue 1) and its only other bad behaviour on \Re(s)\ge 1 would come from a zero of \zeta on that line — this is the single deep input — we get the key holomorphy statement below.

The three ingredients

Newman's proof is a tripod. Each leg is a known result; the cleverness is entirely in how little of each is used and how they combine.

Ingredient (ii) is proved by a wholly elementary counting trick on the middle binomial coefficient \binom{2n}{n} — see Chebyshev's bounds. It is far weaker than the theorem we are proving (it does not say \theta \sim x, only that \theta/x stays bounded), yet it is exactly enough. The third leg is the analytic heart.

Ingredient (iii): Newman's analytic theorem

This is a Tauberian statement in the spirit of the Tauberian theorems: the transform g being finite at the boundary point z=0 is an "Abelian" fact, and the theorem upgrades it — with the side condition that f is bounded — into genuine convergence of the raw integral \int_0^\infty f\,dt. That upgrade normally fails; boundedness is the Tauberian hypothesis that saves it. Newman's contribution was a proof of this theorem in barely half a page, using nothing but Cauchy's integral formula and a shrewd choice of integrand.

Bolting them together

Now apply the analytic theorem to the specially chosen function

By Chebyshev's bound (ii), \theta(e^t) = O(e^t), so f is bounded — the one hypothesis the theorem demands. Its Laplace transform unwinds, after the substitution x = e^{t}, into

Here the two pieces come straight from the Mellin formula (\int_1^\infty \theta(x)x^{-z-2}dx = \Phi(z+1)/(z+1)) and the elementary \int_1^\infty x^{-z-1}dx = 1/z. The apparent pole at z=0 is a mirage: rewrite

and the bracket is holomorphic on \Re(z) \ge 0 precisely by ingredient (i). So g extends holomorphically past the boundary, the analytic theorem fires, and we conclude that

That single convergent integral is the whole harvest of the complex analysis. Everything after it is real, elementary, and short.

Worked example: convergence forces θ(x) ∼ x

Here is the beautiful last step — a "Tauberian squeeze" that uses only that \theta is non-decreasing. We show the convergence of \int_1^\infty \frac{\theta(x)-x}{x^2}\,dx forces \theta(x)/x \to 1.

Suppose the upper side fails. Then there is some fixed \lambda > 1 and arbitrarily large x with \theta(x) \ge \lambda x. Because \theta only ever increases, for every t in the interval [x,\,\lambda x] we have \theta(t) \ge \theta(x) \ge \lambda x. Estimate the slice of the integral over that interval, then substitute t = u x:

The right-hand integral is a positive constant that does not depend on x:

But if \int_1^\infty \frac{\theta(x)-x}{x^2}dx converges, then by the Cauchy criterion the integral over [x,\lambda x] must tend to 0 as x\to\infty. A fixed positive lower bound recurring at arbitrarily large x contradicts that. So \theta(x) \ge \lambda x can hold only finitely often: \limsup \theta(x)/x \le 1.

The lower side is the mirror image. If \theta(x) \le \lambda x with \lambda < 1 for arbitrarily large x, integrate over [\lambda x,\, x] (again using monotonicity, now downward) to get a fixed negative constant \int_{\lambda}^{1}\frac{\lambda-u}{u^2}\,du < 0, contradicting convergence the same way. Hence \liminf \theta(x)/x \ge 1. Squeezing the two bounds:

which is the Prime Number Theorem. Notice that monotonicity of \theta is doing all the work in this step — it is the Tauberian side-condition that turns a convergent integral back into a pointwise asymptotic.

Newman's contour (the half-page trick)

The proof of the analytic theorem is where Newman's ingenuity shows. To recover \int_0^T f\,dt \to g(0) he integrates g(z)\,e^{zT} not around a plain circle but around the boundary C of the region \{\,|z| \le R,\ \Re(z) \ge -\delta\,\} — a big circle sliced off by a vertical line just to the left of the imaginary axis (legitimate because g is holomorphic a little past \Re(z)=0). The real stroke of genius is the correcting kernel he multiplies in,

whose extra factor \bigl(\tfrac1z + \tfrac{z}{R^2}\bigr) equals \tfrac{2\Re(z)}{R^2} on |z|=R. That is tiny on the far right of the circle and it kills the contribution of the large arc, so the whole estimate collapses to give \int_0^T f\,dt \to g(0) as T\to\infty with R chosen large. The figure shows the contour.

No zero-free region ever appears — only the fact that g (hence \Phi, hence \zeta) is holomorphic a hair's breadth past the boundary, which is exactly ingredient (i).